f (x) = 3x2 n = 2 So, dy/dx = nf(x)[f(x)]n-1 = 2(3x2)(x3 - 1)1 = 6x2(x3 - 1)

1. MATHS WORKSHOPS
Differentiation
Business School

2. Outline
The theory of differentiation
How differentiation is done in practice
Application: Finding Maxima and Minima
Conclusion

3. Differentiation Rules Application Conclusion
Outline
The theory of differentiation
How differentiation is done in practice
Application: Finding Maxima and Minima
Conclusion

4. Differentiation Rules Application Conclusion
Differentiation
• A useful way to explore the properties of a function is to find
the derivative.
Definition (Derivative)
The derivative is a measure of how a function changes as its input
changes. More
• The derivative of a function at a chosen input value describes
the best linear approximation of the function near that input
value.
• For single variable functions, f(x), the derivative at a point
equals the slope of the tangent line to the graph of the
function at that point.
• The process of finding a derivative is called differentiation.

5. Differentiation Rules Application Conclusion
Tangent
Definition (Tangent)
The tangent to a curve at a given point is the straight line that
“just touches” the curve at that point. More
Example
x
f(x)

6. Differentiation Rules Application Conclusion
Formal Statement of Differentiation
Definition (Derivative at the point x = a)
d
dx
f(a) = f (a) = lim
h→0
f(a + h) − f(a)
h
This can be explained graphically:
x
f(x)
a
f(a)
a + h1
f(a + h1)
f(a + h1) − f(a)
h1

7. Differentiation Rules Application Conclusion
Formal Statement of Differentiation
Definition (Derivative at the point x = a)
d
dx
f(a) = f (a) = lim
h→0
f(a + h) − f(a)
h
This can be explained graphically:
x
f(x)
a
f(a)
a + h2
f(a + h2)
f(a + h2) − f(a)
h2

8. Differentiation Rules Application Conclusion
Formal Statement of Differentiation
Definition (Derivative at the point x = a)
d
dx
f(a) = f (a) = lim
h→0
f(a + h) − f(a)
h
≈
∆y
∆x
=
rise
run
This can be explained graphically:
x
f(x)
a
f(a)
f (a) =
∆y
∆x
=
rise
run

9. Differentiation Rules Application Conclusion
Example using the technical definition
Example (Differentiate f(x) = x2
by first principles)
Using the definition on the previous slide:
d
dx
f(x) = f (x) = lim
h→0
f(x + h) − f(x)
h
= lim
h→0
(x + h)2 − x2
h
= lim
h→0
x2 + 2xh + h2 − x2
h
= lim
h→0
h(2x + h)
h
= lim
h→0
(2x + h)
= 2x

10. Differentiation Rules Application Conclusion
Notation
• Given some function of x, y = f(x), the following expressions
are equivalent:
dy
dx
=
d
dx
y =
d
dx
f(x) = f (x).
• We read
dy
dx
as “the derivative of y with respect to x.”
• We can differentiate with respect to whatever variable we’d
like. For example if y is a function of u, y = f(u), we can
differentiate y with respect to u:
dy
du
= f (u).
• We read f (x) as f prime x. This notation is often used for
convenience when there is no ambiguity about what we are
differentiating with respect to.

11. Differentiation Rules Application Conclusion
Outline
The theory of differentiation
How differentiation is done in practice
Application: Finding Maxima and Minima
Conclusion

12. Differentiation Rules Application Conclusion
Differentiation in practice
Rarely do people differentiate by “first principles”, i.e. using the
definition. Instead, we use some simple rules: More
Function: f(x) = y Derivative f (x) = dy
dx
xn
nxn−1
axn
anxn−1
a (some constant) 0
log(x) x−1 = 1
x
ex ex
Example (f(x) = x2
)
Using the above rules,
f (x) =
d
dx
f(x) = 2x2−1
= 2x1
= 2x.

13. Differentiation Rules Application Conclusion
More complicated example
Example (f(x) = 2x4
+ 5x3
)
f (x) = 2 × 4x4−1
+ 5 × 3x3−1
= 8x3
+ 15x2
Example (Your turn: f(x) = 9x2
+ x3
)
f (x) =
=
=

14. Differentiation Rules Application Conclusion
More complicated example
Example (f(x) = 2x4
+ 5x3
)
f (x) = 2 × 4x4−1
+ 5 × 3x3−1
= 8x3
+ 15x2
Example (Your turn: f(x) = 9x2
+ x3
)
f (x) = 9 × 2x2−1
+ 3x3−1
=
=

15. Differentiation Rules Application Conclusion
More complicated example
Example (f(x) = 2x4
+ 5x3
)
f (x) = 2 × 4x4−1
+ 5 × 3x3−1
= 8x3
+ 15x2
Example (Your turn: f(x) = 9x2
+ x3
)
f (x) = 9 × 2x2−1
+ 3x3−1
= 18x + 3x2
= 3x(6 + x)

16. Differentiation Rules Application Conclusion
Differentiating Exponential Functions
• The exponential function is unique in that it is equal to its
derivative:
d
dx
ex
= ex
• The exponential function is sometimes written as:
ef(x)
= exp{f(x)}.
• In this definition, a function of x, f(x), is exponentiated.
• The rule for finding a derivative of this type is:
d
dx
exp{f(x)} = f (x) exp{f(x)}.

17. Differentiation Rules Application Conclusion
Differentiating Exponential Functions
Example (g(x) = exp{2x})
1. Rewrite as g(x) = exp{f(x)} where f(x) = 2x.
2. Find, f (x) = 2x1−1 = 2x0 = 2.
3. Thus, g (x) = f (x) exp{f(x)} = 2 exp{2x}.

18. Differentiation Rules Application Conclusion
Differentiating Exponential Functions
Example (g(x) = exp{2x})
1. Rewrite as g(x) = exp{f(x)} where f(x) = 2x.
2. Find, f (x) = 2x1−1 = 2x0 = 2.
3. Thus, g (x) = f (x) exp{f(x)} = 2 exp{2x}.
Example (Your turn: h(x) = exp{3x + 1})
1. Rewrite as h(x) = exp{f(x)} where f(x) = .
2. Find f (x) = .
3. Thus, h (x) = f (x) exp{f(x)} = .

19. Differentiation Rules Application Conclusion
Differentiating Exponential Functions
Example (g(x) = exp{2x})
1. Rewrite as g(x) = exp{f(x)} where f(x) = 2x.
2. Find, f (x) = 2x1−1 = 2x0 = 2.
3. Thus, g (x) = f (x) exp{f(x)} = 2 exp{2x}.
Example (Your turn: h(x) = exp{3x + 1})
1. Rewrite as h(x) = exp{f(x)} where f(x) = 3x + 1.
2. Find f (x) = .
3. Thus, h (x) = f (x) exp{f(x)} = .

20. Differentiation Rules Application Conclusion
Differentiating Exponential Functions
Example (g(x) = exp{2x})
1. Rewrite as g(x) = exp{f(x)} where f(x) = 2x.
2. Find, f (x) = 2x1−1 = 2x0 = 2.
3. Thus, g (x) = f (x) exp{f(x)} = 2 exp{2x}.
Example (Your turn: h(x) = exp{3x + 1})
1. Rewrite as h(x) = exp{f(x)} where f(x) = 3x + 1.
2. Find f (x) = 3x1−1 + 0 = 3.
3. Thus, h (x) = f (x) exp{f(x)} = .

21. Differentiation Rules Application Conclusion
Differentiating Exponential Functions
Example (g(x) = exp{2x})
1. Rewrite as g(x) = exp{f(x)} where f(x) = 2x.
2. Find, f (x) = 2x1−1 = 2x0 = 2.
3. Thus, g (x) = f (x) exp{f(x)} = 2 exp{2x}.
Example (Your turn: h(x) = exp{3x + 1})
1. Rewrite as h(x) = exp{f(x)} where f(x) = 3x + 1.
2. Find f (x) = 3x1−1 + 0 = 3.
3. Thus, h (x) = f (x) exp{f(x)} = 3 exp{3x + 1}.

22. Differentiation Rules Application Conclusion
Differentiating Logarithmic Functions
• As with the exponential function, there are some special rules
for differentiating logarithmic (or log) functions.
• Simple case:
d
dx
log(x) =
1
x
• General case:
d
dx
log(f(x)) =
f (x)
f(x)
Example (g(x) = log(2x2
+ x))
1. Rewrite as g(x) = log(f(x)) where f(x) = 2x2 + x.
2. Find f (x) = 4x + 1.
3. Thus, g (x) =
f (x)
f(x)
=
4x + 1
2x2 + x
.

23. Differentiation Rules Application Conclusion
Differentiating Logarithmic Functions
Example (Your turn: y = log(10x2
− 3x2
))
1. Rewrite as y = log{f(x)} where f(x) = .
2. Find f (x) = .
3. Thus,
dy
dx
=
d
dx
log(10x2
− 3x2
) =
f (x)
f(x)
=
=
= .

24. Differentiation Rules Application Conclusion
Differentiating Logarithmic Functions
Example (Your turn: y = log(10x2
− 3x2
))
1. Rewrite as y = log{f(x)} where f(x) = 10x2 − 3x3.
2. Find f (x) = .
3. Thus,
dy
dx
=
d
dx
log(10x2
− 3x2
) =
f (x)
f(x)
=
=
= .

25. Differentiation Rules Application Conclusion
Differentiating Logarithmic Functions
Example (Your turn: y = log(10x2
− 3x2
))
1. Rewrite as y = log{f(x)} where f(x) = 10x2 − 3x3.
2. Find f (x) = 10 × 2x − 3 × 3x2 = 20x − 9x2 = x(20 − 9x).
3. Thus,
dy
dx
=
d
dx
log(10x2
− 3x2
) =
f (x)
f(x)
=
=
= .

26. Differentiation Rules Application Conclusion
Differentiating Logarithmic Functions
Example (Your turn: y = log(10x2
− 3x2
))
1. Rewrite as y = log{f(x)} where f(x) = 10x2 − 3x3.
2. Find f (x) = 10 × 2x − 3 × 3x2 = 20x − 9x2 = x(20 − 9x).
3. Thus,
dy
dx
=
d
dx
log(10x2
− 3x2
) =
f (x)
f(x)
=
x(20 − 9x)
10x2 − 3x3
=
x(20 − 9x)
x2(10 − 3x)
=
20 − 9x
x(10 − 3x)
.

27. Differentiation Rules Application Conclusion
Other useful differentiation rules
Definition (Specialised Chain Rule)
Let y = [f(x)]n,
dy
dx
= nf (x) [f(x)]n−1
• This is a special case of the chain rule More
Example (y = (x2
+ 2)3
)
Here y = [f(x)]n = (x2 + 2)3 so
• f(x) = x2 + 2 =⇒ f (x) =
d
dx
(x2 + 2) = 2x
• n = 3
• dy
dx = nf (x) [f(x)]n−1
= 3 × 2x × (x2 + 2)3−1 = 6x(x2 + 2)2

28. Differentiation Rules Application Conclusion
Other useful differentiation rules
Definition (Product Rule)
Let y = uv where u and v are functions of x,
dy
dx
= u ·
dv
dx
+ v ·
du
dx
= uv + vu More
Example (y = x3
log(x))
Here u = x3; v = log(x); v =
dv
dx
=
1
x
; u =
du
dx
= 3x2 so
dy
dx
= u × v + v × u
= x3
×
1
x
+ log(x) × 3x2
= x2
+ 3x2
log(x)
= x2
(1 + 3 log(x)).

29. Differentiation Rules Application Conclusion
Your turn. . .
Example (Your turn y = (x3
− 1)2
)
Think of our function as y = [f(x)]n we have in this particular
case,
• f(x) =
• f (x) =
• n =
So,
dy
dx
= nf (x) [f(x)]n−1
=
=

30. Differentiation Rules Application Conclusion
Your turn. . .
Example (Your turn y = (x3
− 1)2
)
Think of our function as y = [f(x)]n we have in this particular
case,
• f(x) = x3 − 1
• f (x) =
• n =
So,
dy
dx
= nf (x) [f(x)]n−1
=
=

31. Differentiation Rules Application Conclusion
Your turn. . .
Example (Your turn y = (x3
− 1)2
)
Think of our function as y = [f(x)]n we have in this particular
case,
• f(x) = x3 − 1
• f (x) =
d
dx
(x3 − 1) = 3x2
• n =
So,
dy
dx
= nf (x) [f(x)]n−1
=
=

32. Differentiation Rules Application Conclusion
Your turn. . .
Example (Your turn y = (x3
− 1)2
)
Think of our function as y = [f(x)]n we have in this particular
case,
• f(x) = x3 − 1
• f (x) =
d
dx
(x3 − 1) = 3x2
• n = 2
So,
dy
dx
= nf (x) [f(x)]n−1
=
=

33. Differentiation Rules Application Conclusion
Your turn. . .
Example (Your turn y = (x3
− 1)2
)
Think of our function as y = [f(x)]n we have in this particular
case,
• f(x) = x3 − 1
• f (x) =
d
dx
(x3 − 1) = 3x2
• n = 2
So,
dy
dx
= nf (x) [f(x)]n−1
= 2 × 3x2
× (x3
− 1)2−1
= 6x2
(x3
− 1)

34. Differentiation Rules Application Conclusion
Your turn. . .
Example (Your turn y = 2x−2
log(x − 1))
We’re differentiating a product so think of the function as y = uv.
• Let u = which means u =
du
dx
=
• Let v = which means v =
dv
dx
=
So,
dy
dx
= uv + vu =
=
Note that you can always check you differentiation using
WolframAlpha: d/dx 2x^(-2)*log(x-1) More

35. Differentiation Rules Application Conclusion
Your turn. . .
Example (Your turn y = 2x−2
log(x − 1))
We’re differentiating a product so think of the function as y = uv.
• Let u = 2x−2 which means u =
du
dx
=
• Let v = which means v =
dv
dx
=
So,
dy
dx
= uv + vu =
=
Note that you can always check you differentiation using
WolframAlpha: d/dx 2x^(-2)*log(x-1) More

36. Differentiation Rules Application Conclusion
Your turn. . .
Example (Your turn y = 2x−2
log(x − 1))
We’re differentiating a product so think of the function as y = uv.
• Let u = 2x−2 which means u =
du
dx
= − 4x−3
• Let v = which means v =
dv
dx
=
So,
dy
dx
= uv + vu =
=
Note that you can always check you differentiation using
WolframAlpha: d/dx 2x^(-2)*log(x-1) More

37. Differentiation Rules Application Conclusion
Your turn. . .
Example (Your turn y = 2x−2
log(x − 1))
We’re differentiating a product so think of the function as y = uv.
• Let u = 2x−2 which means u =
du
dx
= − 4x−3
• Let v = log(x − 1) which means v =
dv
dx
=
So,
dy
dx
= uv + vu =
=
Note that you can always check you differentiation using
WolframAlpha: d/dx 2x^(-2)*log(x-1) More

38. Differentiation Rules Application Conclusion
Your turn. . .
Example (Your turn y = 2x−2
log(x − 1))
We’re differentiating a product so think of the function as y = uv.
• Let u = 2x−2 which means u =
du
dx
= − 4x−3
• Let v = log(x − 1) which means v =
dv
dx
=
1
x − 1
So,
dy
dx
= uv + vu =
=
Note that you can always check you differentiation using
WolframAlpha: d/dx 2x^(-2)*log(x-1) More

39. Differentiation Rules Application Conclusion
Your turn. . .
Example (Your turn y = 2x−2
log(x − 1))
We’re differentiating a product so think of the function as y = uv.
• Let u = 2x−2 which means u =
du
dx
= − 4x−3
• Let v = log(x − 1) which means v =
dv
dx
=
1
x − 1
So,
dy
dx
= uv + vu = 2x−2 1
x − 1
+ (−4x−3
) log(x − 1)
=
Note that you can always check you differentiation using
WolframAlpha: d/dx 2x^(-2)*log(x-1) More

40. Differentiation Rules Application Conclusion
Your turn. . .
Example (Your turn y = 2x−2
log(x − 1))
We’re differentiating a product so think of the function as y = uv.
• Let u = 2x−2 which means u =
du
dx
= − 4x−3
• Let v = log(x − 1) which means v =
dv
dx
=
1
x − 1
So,
dy
dx
= uv + vu = 2x−2 1
x − 1
+ (−4x−3
) log(x − 1)
=
2
x2(x − 1)
− 4x−3
log(x − 1)
Note that you can always check you differentiation using
WolframAlpha: d/dx 2x^(-2)*log(x-1) More

41. Differentiation Rules Application Conclusion
Other useful differentiation rules
Definition (Quotient Rule)
Let y =
u
v
where u and v are functions of x then
dy
dx
=
v ·
du
dx
− u ·
dv
dx
v2
=
vu − uv
v2
More
Example (y =
x4
ex
)
Let u = x4 and v = ex; then u =
du
dx
= 4x3 and v =
dv
dx
= ex,
dy
dx
=
vu − uv
v2
=
ex · 4x3 − x4 · ex
(ex)2
=
x3(4 − 1)
ex
.

42. Differentiation Rules Application Conclusion
Outline
The theory of differentiation
How differentiation is done in practice
Application: Finding Maxima and Minima
Conclusion

43. Differentiation Rules Application Conclusion
Finding Maxima and Minima
The most common use of differentiation is to find the maximum
and minimum values of functions.
Key Idea
“Stationary points” occur when the derivative equals zero,
f (x) = 0, i.e. the tangent line is a horizontal line. More
x
f(x)

44. Differentiation Rules Application Conclusion
Finding Maxima and Minima
To determine if a stationary point is a maximum, minimum or
neither, we find the second order derivatives.
Definition (Second order derivative)
The second order derivative of a function, f(x), is found by taking
the derivative of the first order derivative:
f (x) =
d
dx
f (x) =
d
dx
d
dx
f(x) .
• If f (x) < 0, the stationary point at x is a maximum.
• If f (x) > 0, the stationary point at x is a minimum.
• If f (x) = 0, the nature of the stationary point must be
determined by way of other means, often by noting a sign
change around that point.

45. Differentiation Rules Application Conclusion
Finding Turning Points
Example (f(x) = 2x4
+ 5x3
)
We found previously that f (x) = 8x3 + 15x2 = x2(8x + 15).
• To find the turning points, we set f (x) = 0:
x2
(8x + 15) = 0
• This occurs when either:
• x2
= 0 =⇒ x = 0 (this is a point of inflection) More
• 8x + 15 = 0 =⇒ x = −
15
8
= −1.875 (a minimum) More
• To determine whether these are turning points are maxima,
minima or neither we find the second order derivative:
f (x) =
d
dx
f (x) =
d
dx
8x3
+ 15x2
= 24x2
+ 30x

46. Differentiation Rules Application Conclusion
Finding Turning Points
Example (f(x) = 2x4
+ 5x3
continued)
• The second order derivative is: f (x) = 24x2 + 30x
• We need to evaluate f (x) at the values of x we identified as
turning points:
• f (0) = 0 =⇒ a point of inflection More
• f (−1.875) = 24 × (−1.875)2
− 30 × 1.875 = 28.125 > 0
=⇒ a minimum More
x
f(x)
-1.875

47. Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima. . .
Given f(x) = 9x2 + x3, previously you found f (x) = 18x + 3x2.
• To find the turning points set f (x) = 0:
• This occurs when either:
• or
•
• Second order derivative: f (x) = . Evaluate this at
the possible turning points:
•
•

48. Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima. . .
Given f(x) = 9x2 + x3, previously you found f (x) = 18x + 3x2.
• To find the turning points set f (x) = 0:
3x(6 + x) = 0
• This occurs when either:
• 3x = 0 =⇒ x = 0 or
• 6 + x = 0 =⇒ x = −6
• Second order derivative: f (x) = . Evaluate this at
the possible turning points:
•
•

49. Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima. . .
Given f(x) = 9x2 + x3, previously you found f (x) = 18x + 3x2.
• To find the turning points set f (x) = 0:
3x(6 + x) = 0
• This occurs when either:
• 3x = 0 =⇒ x = 0 or
• 6 + x = 0 =⇒ x = −6
• Second order derivative: f (x) = 18 + 6x. Evaluate this at
the possible turning points:
•
•

50. Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima. . .
Given f(x) = 9x2 + x3, previously you found f (x) = 18x + 3x2.
• To find the turning points set f (x) = 0:
3x(6 + x) = 0
• This occurs when either:
• 3x = 0 =⇒ x = 0 or
• 6 + x = 0 =⇒ x = −6
• Second order derivative: f (x) = 18 + 6x. Evaluate this at
the possible turning points:
• f (0) = 18 > 0 =⇒
• f (−6) = 18 + 6 × (−6) = −18 < 0 =⇒

51. Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima. . .
Given f(x) = 9x2 + x3, previously you found f (x) = 18x + 3x2.
• To find the turning points set f (x) = 0:
3x(6 + x) = 0
• This occurs when either:
• 3x = 0 =⇒ x = 0 or
• 6 + x = 0 =⇒ x = −6
• Second order derivative: f (x) = 18 + 6x. Evaluate this at
the possible turning points:
• f (0) = 18 > 0 =⇒ a minimum
• f (−6) = 18 + 6 × (−6) = −18 < 0 =⇒ a maximum
x
f(x)
-6

52. Differentiation Rules Application Conclusion
Maximising Utility
• An investor gains what is known as utility from increasing
his/her wealth (think of utility as simply, enjoyment).
• You can define someones utility as a function of wealth.
Example (Your turn: U(w) = 4w − 1
10
w2
)
• Differentiate U with respect to w:
• Set U (w) = 0 to find the critical points:
• w = gives the theoretical level of wealth for this investor
that will maximise their utility.

53. Differentiation Rules Application Conclusion
Maximising Utility
• An investor gains what is known as utility from increasing
his/her wealth (think of utility as simply, enjoyment).
• You can define someones utility as a function of wealth.
Example (Your turn: U(w) = 4w − 1
10
w2
)
• Differentiate U with respect to w: U (w) = 4 −
2
10
w.
• Set U (w) = 0 to find the critical points:
• w = gives the theoretical level of wealth for this investor
that will maximise their utility.

54. Differentiation Rules Application Conclusion
Maximising Utility
• An investor gains what is known as utility from increasing
his/her wealth (think of utility as simply, enjoyment).
• You can define someones utility as a function of wealth.
Example (Your turn: U(w) = 4w − 1
10
w2
)
• Differentiate U with respect to w: U (w) = 4 −
2
10
w.
• Set U (w) = 0 to find the critical points:
2
10
w = 4 =⇒ w = 20.
• w = gives the theoretical level of wealth for this investor
that will maximise their utility.

55. Differentiation Rules Application Conclusion
Maximising Utility
• An investor gains what is known as utility from increasing
his/her wealth (think of utility as simply, enjoyment).
• You can define someones utility as a function of wealth.
Example (Your turn: U(w) = 4w − 1
10
w2
)
• Differentiate U with respect to w: U (w) = 4 −
2
10
w.
• Set U (w) = 0 to find the critical points:
2
10
w = 4 =⇒ w = 20.
• w = 20 gives the theoretical level of wealth for this investor
that will maximise their utility.

56. Differentiation Rules Application Conclusion
Applications in Business
• The ubiquitous Cobb-Douglas production function uses
exponentials and logs More
• The formal interpretation of regression coefficients in
econometrics requires differentiation More
• Differentiation to finding maxima is used for constrained
optimisation in operations management More
• Marginal benefits and marginal costs can be derived using
differentiation More

57. Differentiation Rules Application Conclusion
Summary
• Functions, log and exponential functions
• Differentiation tells us about the behaviour of the function
• The derivative of a single variable function is the tangent
• The derivative can be interpreted as the “rate of change” of
the function
• Chain rule, product rule, quotient rule
• Finding maxima and minima
• Applications in Business

58. Differentiation Rules Application Conclusion
Summary of Differentiation Identities
Function Derivative
f(x) = axn
f (x) = anxn−1
f(x) = a (some constant) f (x) = 0
f(x) = exp{g(x)} f (x) = g (x) · exp{g(x)}
y = log{f(x)}
dy
dx
=
f (x)
f(x)
y = f(u), u = g(x)
dy
dx
=
dy
du
·
du
dx
y = uv, u = g(x), v = h(x)
dy
dx
= u ·
dv
dx
+ v ·
du
dx
y =
u
v
, u = g(x), v = h(x) dy
dx
=
v ·
du
dx
− u ·
dv
dx
v2
More

59. Differentiation Rules Application Conclusion
Additional Resources
• Test your knowledge at the University of Sydney Business
School MathQuiz:
http://quiz.econ.usyd.edu.au/mathquiz
• Additional resources on the Maths in Business website
sydney.edu.au/business/learning/students/maths
• The University of Sydney Mathematics Learning Centre has a
number of additional resources:
• Maths Learning Centre algebra workshop notes More
• Other Maths Learning Centre Resources More
• The Department of Mathematical Sciences and the
Mathematics Learning Support Centre at Loughborough
University have a fantastic website full of resources. More
• There’s also tonnes of theory, worked questions and additional
practice questions online. All you need to do is Google the
topic you need more practice with! More

60. Differentiation Rules Application Conclusion
Acknowledgements
• Presenters and content contributors: Garth Tarr, Edward
Deng, Donna Zhou, Justin Wang, Fayzan Bahktiar, Priyanka
Goonetilleke.
• Mathematics Workshops Project Manager Jessica Morr from
the Learning and Teaching in Business.
• Valuable comments, feedback and support from Erick Li and
Michele Scoufis.
• Questions, comments, feedback? Let us know at
business.maths@sydney.edu.au